For the circuit shown 11-1 ??, R1 = 5000 ?,RL= 5000 ? VDDP-12 V and VDDN-12...
For the circuit shown 11-20 μΑ, 12-2 μ A, R,-5000 Ω. RL-200 Ω, VDDP-12 V and VDDN-12 V and the OPAMP is ideal and you notice there is negative feedback. VoUT due to I1 mV OUT due to I,- mV VoUT due to both I1 and 12 mV R1 (121 12 RL (RL) The relative tolerance for this problem is 1 %.
For the circuit shown, VIN 2 mV RI R3-60 k2, R2-R 60 k2, VDDP 12 V and VDDN 12 V and the OPAMP is ideal and you notice there is negative feedback. VOUT R1 R2 R3 R4 iR11 (R2) iR1) fR21 - IN V3 (VIN) The relative tolerance for this problem is 1 % Get help: Written Example License Question 1. Points possible: 10 Unlimited attempts For the circuit shown. negative feedback. Vr.-2 mV RI-R3-2 kQ. R2_R4-120 kQ. VDDp-12 V...
Help! For the circuit shown, negative feedback. Vn = 1 mV R,-R,-60 kQ, R =R, 120 ㏀ VDD 9 V and VDDN 9 V and the OPAMP s ideal and you notice there is OUT =10.04 R1 R2 R3 R4 R2) IR2) V3 U1 U2 (VIN) The relative tolerance for this problem is 1 %.
For the circuit shown below, design the values of R1 and R2 that will cause Vout-6 Vif VIv-9 V. If the resistors you chose have a tolerance of 1 %, what is the maximum and minimum values of Vout? Design R3 so that the LED will be properly lit, if IF=10 mA, VF-3 V and Vout-6 V Design equations from the data sheet: VOUT-1.25%(1+R2 R1) U1 OUT ININLT OUT R3 R1 V1 (R3) R1) ADJ LT317A D1 IN) R2 (R2)...
Calculate Vout if Avoc=100000, VINP=0, VINN=-1.0E-5, VDD+=12, VDD-=-12, R1=1kN, R2=2kN, Vout= Preview V. in- in+ N.U1a out 14 3R1 {VINN} {R1} >R2 <{R2} {VINP} The relative tolerance for this problem is 1 %.
Assuming an ideal op-amp in the following circuit, find output voltage, Vo if R1= 2 K2, R2=8 K12, R3=3.8 KS2, R4=6 KI2, R5=15 KS2, R6=3.8 KN, RL=9.8 K12, V1=1V, 12=0.5 mA and V3=2.2 V. } R6 R1 w R5 w + Vo + } RL 12 R2 V1 R3 R4 + +1 V3 Using the above circuit, but consider the following component values: R1= 2 KN R2=8 K2, R3=4.1 K12, R4=6 KI2, R5=17.0 K12, R6=15 KI, RL=10 KI, V1=1V, 12=0.5mA...
In the circuit shown in (Figure 1), E = 29.0 V, R1 = 6.00 12, R3 = 12.0 12, and R2 can vary between 3.00 N and 24.0 12. Part A For what value of R2 is the power dissipated by heating element Rị the greatest? Express your answer with the appropriate units. НА ? R2 Value Units Submit Request Answer Part B Calculate the magnitude of the greatest power. Express your answer with the appropriate units. μΑ PE Value...
For the driven-right-leg system below along with its equivalent circuit, assume 4. R,-5M0(non-ideal input impedance of op-amp) db Rx Auxiliary op amp RL db R1 Vcm_ Vi R5 2 RF R4 Rx o v RG 0 7 RA Auxiliary Amplifier O VREF 2 R2 What are the benefits of using the driven-right-leg circuit? What is the value of the common-mode voltage if resistor RRL is tied to ground instead of the auxiliary amplifier? e. For the driven-right-leg system below along...
NONINVERTING AMPLIFIER For the circuit below, assume: Vs = 8 Vpp, R1 = 1600 2, R2 = 3800 and R3 = 5200 . Find the values of the four currents, i1, i2, i3 and iout. (Format all answers to two decimal points): i 1 = mA, 12 = mA, i3 = mA and jout = mA. iout Op Amp Vout R2 We have negative feedback, so the golden rules tell us: 1) Iin = 0 + vt = Vs. 2)...
R1 R2 AAA 4700 1000 12 LED1 R4 2200 Figure 1 Part 4 Application of Thevenin's Equivalent Circuit The LED in Figure 1 is replaced with a resistor RL Predict the voltage across Rl and the current in RL for RL = 4.7k12, 10k12, and 47 12 respectively. Enter your values in Table 4. Measure Rx and the voltage across and current in it for the three values of RL, and enter your results in Table 4. Workings: Table 4...